$ 2.0 \; L $ perchloric acid $ HClO_4 $ with concentration $ 20.1 \frac{g}{L} $ are mixed with $ 5.0 \; L $ of the same acid with concentration $ 0.20 \; M $. To calculate the pH, it is first necessary to calculate the number of moles of acid in each solution:
$n_1$ $=$ $\frac{2.0\cdot 20.1}{100.05}$ $=$ $0.40\; mol $ $n_2$ $=$ $5.0\cdot 0.20$ $=$ $1.0\;mol$
Then we calculate the final molarity of the acid:
$[HClO_4]$ $=$ $\frac{1.0+0.40}{7.0}$ $=$ $0.20\frac{mol}{L}$
Finally, we calculate the pH:
$pH$ $=$ $-log[HClO_4]$ $=$ $-log0.020$ $=$ $1.70$
